Question

A random sample of 200 New York State voters included 88 Republicans, while a random sample of 300 California voters produced 141 Republicans. Which of the following represents the $95\%$confidence interval that should be used to estimate the true difference in the proportions of Republicans in New York State and California?

(a) $(0.44-0.47)\pm 1.96\frac{(0.44)(0.56)+(0.47)(0.53)}{\sqrt{200+300}}$

(b) $(0.44-0.47)\pm 1.96\frac{(0.44)(0.56)}{\sqrt{200}}+\frac{(0.47)(0.53)}{\sqrt{300}}$

(c) $(0.44-0.47)\pm 1.96\sqrt{\frac{(0.44)(0.56)}{200}+\frac{(0.47)(0.53)}{300}}$

(d)$(0.44-0.47)\pm 1.96\sqrt{\frac{(0.44)(0.56)+(0.47)(0.53)}{200+300}}$

(e) $(0.44-0.47)\pm 1.96\sqrt{\left(045\right)(0.55)\left(\frac{1}{200}+\frac{1}{300}\right)}$

Short answer

$\text{(c)}(0.44-0.47)-1.96·\sqrt{\frac{0.44(0.56)}{200}+\frac{0.47(0.53)}{300}}$

Step-by-step solution

Given Information

$n_1=200$

$x_1=88$

$n_2=300$

$x_2=141$

Explanation

The formula confidence interval of difference of proportions:

$\left({\hat{p}}_1-{\hat{p}}_2\right)\pm z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

The sample proportion is calculated by dividing the number of successes by the sample size.:

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{88}{200} \\ =0.44 \end{gathered}$$

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{141}{300} \\ =0.47 \end{gathered}$$

Determine the critical value using the table $z$:

localid="1650359734870" $$\begin{gathered} z_{\alpha /2}=z_{0.025} \\ =1.96 \end{gathered}$$

The confidence interval is then:

localid="1654604168020" $(0.44-0.47)\pm 1.96·\sqrt{\frac{0.44(1-0.44)}{200}+\frac{0.47(1-0.47)}{300}}$

localid="1654604178645" $=(0.44-0.47)\pm 1.96·\sqrt{\frac{0.44(0.56)}{200}+\frac{0.47(0.53)}{300}}$